如何在Python中的“ with”语句下调用整个函数？

Question

I have the following sample function:

def Run(self):
    with self._resource() as r:
        # a lot of code uses |r|
        pass
    # end of 'with' statement
# end of function body

I don't want to lose the visual space of the whole function body because of additional indent inside with statement.

Also I don't want to call the _resource() outside the class scope - it breaks encapsulation in some ways, ie this is not a good way:

with obj._resource() as r:
    obj.Run(r)

Is there any pretty way to run the same code without losing visual space?

Answer 1

If you have only this function to deal with, it's pretty simple:

class Foo(object):
    def Run(self):
        with self._resource() as r:
            return self._RunWithResource(r)

    def _RunWithResource(self, r):
        # ...

If you want to repeat the pattern, a decorator might help. More or less:

from functools import wraps
def with_resource(f):
    @wraps
    def wrapper(self, *a, **kw):
        with self._resource() as r:
            return f(self, r, *a, **kw)
    return wrapper

class Foo(object):
    @with_resource
    def Run(self, r):
        # ...